Question

A man wishes to vacuum his car with a canister vacuum cleaner marked 535 W at 120. V. The car

is parked far from the building, so he uses an extension cord 15.0 m long to plug the cleaner into a

120. - V source. Assume the cleaner has constant resistance. (a) If the resistance of each of the              

two conductors of the extension cord is 0.900 , what is the actual power delivered to the                             

cleaner? (b) If, instead, the power is to be at least 525 W, what must be the diameter of each of          

two identical copper conductors in the cord the young man buys? (c) Repeat part (b) if the power

is to be at least 532 W. Suggestion: A symbolic solution can simplify the calculations.

Answer 1

a.

Resistance of cleaner R_C = Voltage²/ power = 120² /535 = 26.92 ohms

Resistance of extension cord

R_E = 2 x 0.900 = 1.80 oms

total resistance R = R_C + R_E = 26.92 + 1.80 = 28.72 ohms

Current, I = V /R = 120 /28.92 = 4.18 A

Power delivered to cleaner P_C = I²*R_C = 4.18²*26.92 = 470.10 W

b.

If power is P_C^' = 525 W,

P_C' = V² /R^' => R^' = 120²/525 = 27.43

Resistance of new cord R_E^' = R^' - R_C = 27.43 - 26.92 = 0.51 ohms

Also R_E^' = * L / A = * L / pi *(d²/4)

= resistivity of copper = 1.72 x 10^-8 ohm-m ,

L = Length of cord = 2 *15 = 30 m

d = diameter of wire

0.51 = (1.72 x 10^-8 x 30 x 4) / 3.14 x d²

d = 1.14 x 10^-3 m